A charge of 4μC is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:
(A)1μC and 3μC
(B)2μC and 2μC✓ Correct
(C)0 and 4μC
(D)1.5μC and 2.5μC
Answer: (B)
Solution
Let the charges be q and 4−q. The force F=d2Kq(4−q). To maximize F, differentiate with respect to q: dqdF=d2K(4−2q)=0, giving q=2 μC. Thus the charges should be 2 μC and 2 μC.
JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option